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3C24 Nuclear and Particle Physics Lecture 10 (UCL)


3C24 Nuclear and Particle Physics Lecture 1-10 (UCL), astronomy, astrophysics, cosmology, general relativity, quantum mechanics, physics, university degree, lecture notes, physical sciences

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									7.    STRUCTURE OF NUCLEI                                                                           Finally, using p = hk , gives the result above.] Since every state can contain two fermions of
In section 2 we looked at some of the phenomena of nuclear physics. In this section we return       the same species, we can have (using n = 2 Ú                   dn )
to nuclear physics and discuss some theoretical models that tell us more about the structure of                                                           0
                                                                                                                                           V ( pF ) 3
                                                                                                                                                                                    V ( pF ) 3
                                                                                                                                    N=                         and Z =
7.1 Fermi Gas Model                                                                                                                         3p 2 h 3                                 3p 2 h 3
In this model, the protons and neutrons that make up the nucleus are assumed to comprise two
independent systems of nucleons, each freely moving inside the nuclear volume subject to the        neutrons and protons, respectively. With a nuclear volume
constraints of the Pauli principle. The potential felt by every nucleon is the superposition of
the potentials due to all the other nucleons. In the case of neutrons this is assumed to be a                                                         4        4
                                                                                                                                            V=                     3
                                                                                                                                                        p R 3 = p R0 A
finite-depth square well; for protons, the Coulomb potential modifies this. A sketch of the                                                           3        3
potential wells in both cases is shown in Fig.7.1.
                                                                                                    where experimentally R0 = 1.21fm (as we have seen from electron scattering experiments
                                                                                                    discussed in Section 2) and assuming the depths of the neutron and proton wells to be the
                                                                                                    same, we find for a nucleus with Z = N = A 2 , the Fermi momentum
                                                                                                                                                                            1/ 3
                                                                                                                                      n    p             h Ê 9p ˆ
                                                                                                                                pF = pF = pF =              Á ˜                    ª 250MeV c
                                                                                                                                                         R0 Ë 8 ¯
                                                                                                    Thus the nucleons move freely within the nucleus with large momenta.
                                                                                                    The Fermi energy is
                                                                                                                                                EF =        ª 33MeV
           Fig.7.1 Proton and neutron potentials and states in the Fermi gas model                  The difference B¢ between the top of the well and the Fermi level is constant for most nuclei
                                                                                                    and is just the average binding energy per nucleon B A = 7 - 8MeV . The depth of the
For a given ground state nucleus, the energy levels will fill up from the bottom of the well.       potential and the Fermi energy are to a good approximation independent of the mass number
The energy of the highest level that is completely filled is called the Fermi level of energy E F   A:
and has a momentum pF = 2 ME F , where M is the mass of the nucleon. Within the volume                                                V0 = E F + B¢ ª 40 MeV .
V, the number of states with a momentum between p and p + dp is given by
                                                                                                    Heavy nuclei generally have a surplus of neutrons. Since the Fermi levels of the protons and
                                                                                                    neutrons in a stable nucleus have to be equal (otherwise the nucleus can become more stable
                                                    4p V 2
                                  n ( p) dp = dn =         p dp                                     by b -decay) this implies that the depth of the potential well for the neutron gas has to be
                                                   (2ph) 3                                          deeper than for the proton gas, as shown in Fig.7.1. Protons are therefore on average less
                                                                                                    tightly bound in nuclei than are neutrons.
[Note: This is the so-called density of states factor and may be found as follows. Firstly, solve
the Schrödinger equation for a free particle in a box of side L, with volume V = L3 , subject to    We can use the Fermi gas model to give a theoretical expression for some of the dependence
the boundary conditions that the wave function vanishes on all faces of the cube. This restricts    of the binding energy on the surplus of neutrons, as follows.
the allowed wave numbers k to the values
                                                                                                    First we define the average kinetic energy per nucleon
                                         Êp   p     p ˆ
                                     k = Á n1, n 2 , n 3 ˜
                                         ËL   L     L ¯                                                                                     pF                 2
                                                                                                                               E kin   =
                                                                                                                                           Ú0        E kin p       dp
                                                                                                                                                                             3 pF
                                                                                                                                                                                  ª 20MeV
                                                                                                                                                 pF                          5 2M
where n1 etc are integers. Plotting these in a three dimensional grid gives the number of
                                                                                                                                                Ú0     p 2 dp
modes of standing waves with wave number k having magnitude in the range k to k + dk as
                                                                                                    The total kinetic energy of the nucleus (which is a good approximation to B¢ ) is then
                                                    V k2
                                       f ( k ) dk =      dk
                                                    2p 2                                                                                                                  3
                                                                                                                       E kin ( N , Z ) = N E n + Z E p =
                                                                                                                                                                        10 M
                                                                                                                                                                                    [          p
                                                                                                                                                                             N ( pF ) 2 + Z ( pF ) 2   ]
                                              7.1                                                                                                              7.2
which may be re-expressed as                                                                      be occupied by an electron with spin “up” or “down”, corresponding to the spin-projection
                                                                                                  quantum number
                                               3 h 2 Ê 9p ˆ
                                                                   2/3 È N 5/3 + Z 5/3 ˘                                                  ms = ±1 2
                         E kin ( N , Z ) =         2Á     ˜            Í               ˙
                                             10 M R0 Ë 4 ¯             Í
                                                                       Î     A 2/3     ˙
                                                                                       ˚          Again, both these states will have the same energy. So finally, any energy eigenstate in the
                                                                                                  hydrogen atom is labeled by the quantum numbers ( n, l, ml , ms ) and for any n, there will be
where again we have taken the radii of the proton and neutron wells to be equal. This             n d degenerate energy states, where
expression is for fixed A but varying N and has a minimum at N = Z . Hence the binding
energy gets smaller for N π Z . If we set N = ( A + D) 2 , Z = ( A - D) 2 and expand this                                                     n -1
expression in a power series in D A , where D ∫ N - Z , we obtain                                                                     n d = 2 Â (2l + 1) = 2 n 2
                                                                                                                                              l= 0
                                           3 h 2 Ê 9p ˆ           È    5 (N - Z )2       ˘
                     E kin ( N , Z ) =         2 Á    ˜           ÍA + 9           + ....˙        This high degree of degeneracy can be broken if there is a preferred direction in space, such as
                                         10 M R0 Ë 8 ¯            Î         A            ˚        that supplied by a magnetic field, in which case the energy levels could depend on ml and ms .
                                                                                                  One such interaction is the spin-orbit coupling, which is the interaction between the magnetic
which gives the dependence on the neutron surplus. The first term contributes to the volume       moment of the electron (due to its spin) and the magnetic field due to the motion of the
term in the semi-empirical mass formula (SEMF), while the second describes the correction         nucleus (in the electron rest frame). This leads to corrections to the energy levels called fine
that results from having N π Z . This is the asymmetry energy we have met before and grows        structure.
as the square of the neutron surplus. In practice, to reproduce the actual term in the SEMF
accurately we would have to take into account the change in the potential for N π Z .
                                                                                                  In atomic physics, the fine structure corrections are small (of order a 2 ) and so if we ignore
7.2 Shell Model                                                                                   them for the moment, in hydrogen we would have a system with electron orbits corresponding
                                                                                                  to shells of a given n, with each shell having degenerate sub-shells specified by the orbital
(a) Basic ideas                                                                                   angular momentum l . Going beyond hydrogen, we can introduce the electron-electron
This model is based on the analogous model for the orbital structure of atomic electrons in       Coulomb interaction. This introduces a splitting to any energy level n according to the l
atoms. It some areas it gives more detailed predictions than the Fermi gas model. Firstly, we     value. The degeneracies in ml and ms are unchanged. It is straightforward to see that if a shell
recap the main features of the atomic case.                                                       or sub-shell is filled, then we have
The binding energy of electrons in atoms is due primarily to the central Coulomb potential.                                         Âm   s   = 0 and       Âm   l   =0
This is a complicated problem to solve in general because in a multi-electron atom we have to
take account of not only the Coulomb field of the nucleus, but also the fields of all the other   i.e. there is a strong pairing effect for closed shells. In these cases it can be shown that the
electrons. Analytic solutions are not usually possible. However, many of the general features     Pauli principle implies
of the simplest case of hydrogen carry over to more complicated cases, so it is worth recalling
these. Atomic energy levels are characterised by a quantum number n = 1, 2 , 3, 4 , ....                                          L = S = 0 and J = L + S = 0
In atomic physics, n is called the principal quantum number and is defined so that it             For any atom with a closed shell or a closed sub-shell structure, the electrons are paired off
determines the energy of the system. In nuclear physics we are not dealing with the same          and thus no valence electrons are available. Such atoms are therefore chemically inert. It is
simple coulomb potential, so it would be better to call n the radial node quantum number, as it   straightforward to work out the atomic numbers at which this occurs. These are
still determines the form of the radial part of the wavefunction. For any n there are energy-
degenerate levels with orbital angular momentum quantum numbers given by                                                                Z = 2 ,10 ,18 , 36 , 54
                                           l = 0 ,1, 2 , 3,....,( n - 1)                          For example, the inert gas argon Ar ( Z = 18) has closed shells corresponding to n = 1, 2 and
                                                                                                  closed sub-shells corresponding to n = 3, l = 0 ,1. These values of Z are called the magic
(this restriction follows from the form of the Coulomb potential) and for any value of l there    numbers.
are (2l + 1) sub-states with different values of the projection of orbital angular momentum
along any chosen axis (the magnetic quantum number)                                               (b) Nuclear magic numbers
                                                                                                  In nuclear physics, there is also evidence for magic numbers, i.e. values of Z and N at which
                                  ml = -l , - l + 1 ,..., 0 , 1 ,..., l - 1, l                    the binding is particularly strong. This can been seen from the B A curve of Fig.7.2, where at
                                                                                                  certain values of N and Z the data lie above the SEMF. (The figure only shows results for even
Due to the rotational symmetry of the Coulomb potential, all such sub-states will be              values of the mass number A).
degenerate in energy. Furthermore, since electrons have spin–1/2, each of the states above can
                                                      7.3                                                                                            7.4
                                                                                                  The crucial step in understanding the origin of the magic numbers was taken in 1949 by
                                                                                                  Mayer and Jensen who suggested that by analogy with atomic physics there should also be a
                                                                                                  spin-orbit part, so that the total potential is
                                                                                                                                       Vtot = Vcentral ( r) + Vls ( r) L.S
                                                                                                  where L and S are the orbital and spin angular momentum operators for a single nucleon and
                                                                                                  Vls ( r) is an arbitrary function of the radial coordinate. This form for the total potential is the
                                                                                                  same as used in atomic physics except for the presence of the function Vls ( r) . Once we have
                                                                                                  coupling between L and S then ml and ms are no longer “good” quantum numbers and we
                                                                                                  have to use eigenstates of the total angular momentum vector J, defined by J = L + S .
                                                                                                  Squaring this, we have
                                                                                                                                            J 2 = L2 + S2 + 2 L.S
                                                                                                                                           L.S =   1
                                                                                                                                                   2   (J2 - L2 - S2 )
                                                                                                  and hence the expectation value of L.S , which we write as ls is
                                                                                                                    ls   j ( j + 1) - l(l + 1) - s( s + 1) Ïl 2         for j = l + 1 / 2
                  Fig.7.2 Binding energy per nucleon for even values of A.                                             =                                  =Ì
                               The solid curve is the SEMF.                                                         h2                   2                 Ó- (l + 1) 2 for j = l - 1 / 2
The magic numbers are                                                                             (We are always dealing with a single nucleon, so that s = 1 / 2 .) Thus the splitting between the
                                                                                                  two levels is
                                  N = 2 , 8 , 20 , 28 , 50 , 82 ,126
                                  Z = 2 , 8 , 20 , 28 , 50 , 82                                                                                        2l + 1 2
                                                                                                                                           DE ls =           h Vls
and correspond to one or more closed shells, plus 8 nucleons filling the s and p subshells of a
                                                                                                  Experimentally, it is found that Vls ( r) is negative, which means that the state with j = l + 1 / 2
nuclei with a particular value of n. Nuclei with both N and Z having one of these values are
called doubly magic, and have even greater stability.                                             will have a lower energy than the state with j = l - 1 / 2 . This is opposite to the situation in
                                                                                                  atoms. Also, the splittings are substantial and increase linearly with l . Hence for higher l ,
Shell structure is also suggested by a number of other phenomena. For example: “magic”            level crossings can occur. Namely, for large l , the splitting of any two neighbouring
nuclei have many more stable isotopes than other nuclei; they lack electric dipole moments,       degenerate levels can shift the j = l - 1 / 2 state of the initial lower level to lie above the
which means they are spherical; and neutron capture cross-sections show sharp drops                j = l + 1 / 2 level of the previously higher level.
compared to neighbouring nuclei. However, to proceed further we need to know something
about the effective potential.                                                                    An example of the resulting splittings up to the 1G state is shown in Fig.7.3, where the usual
                                                                                                  atomic spectroscopic notation has been used, i.e. levels are written nl j with
A simple Coulomb potential is clearly not appropriate and we need some form that describes         S , P , D, F , G , .... used for l = 0 , 1, 2 , 3, 4,... In this case a simple finite square well has been
the effective potential of all the other nucleons. Since the nuclear force is short-ranged we     used for the central part of the potential. Magic numbers occur when there are particularly
would expect the potential to follow the form of the density distribution of nucleons in the      large gaps between groups of levels. Note that there is no restriction on the values of l for a
nucleus. So, for example, for very light nuclei this would be Gaussian and the potential could    given n, unlike in the atomic case, because we do not have a Coulomb potential.
be an harmonic oscillator. For heavy nuclei, we have seen that the Fermi distribution fits the
data and the corresponding potential is called the Woods-Saxon form
                                   Vcentral ( r) =
                                                     1 + e ( r - R )/ a
However, although these potentials can be shown to offer an explanation for the lowest magic
numbers, they do not work for the higher ones. This is true of all purely central potentials.
                                                7.5                                                                                                     7.6
                                                                                                       In fact it is found that all even-Z/even-N nuclei have zero nuclear spin. We can therefore make
                                                                                                       the hypothesis that for ground state nuclei, pairs of neutrons and pairs of protons in a given
                                                                                                       sub-shell always couple to give a combined angular momentum of zero, even when the sub-
                                                                                                       shell is not filled. This is called the pairing hypothesis. We can now see why it is the last
                                                                                                       proton and/or last neutron, which determines the net nuclear spin, because these are the only
                                                                                                       ones that may not be paired up. In odd-A nuclides there is only one unpaired nucleon, so we
                                                                                                       can predict precisely what the nuclear spin will be by referring to the filling diagram. For
                                                                                                       even-A odd-Z/odd-N nuclides we will have an unpaired proton and an unpaired neutron. We
                                                                                                       cannot then make a precise prediction about the net spin because of the vectorial way that
                                                                                                       angular momenta combine; all we can say is that the nuclear spin will lie in the range
                                                                                                                  (        )
                                                                                                        j p - jn to j p + jn .
                                                                                                       Predictions can also be made about nuclear parities. Recall: (i) parity is the transformation
                                                                                                       r Æ -r ; (ii) the wavefunction of a single-particle quantum state will contain an angular part
                                                                                                       proportional to the spherical harmonic Ym (q ,f ) , which under the parity transformation
                                                                                                                                           PYm (q ,f ) = (-) l Ym (q ,f )
                                                                                                                                             l                  l
                    Fig.7.3 Energy levels in a single-particle shell model.                            Thus the parity of a single-particle quantum state depends exclusively on the orbital angular
                 The boxed integers correspond to the magic nuclear numbers.
                                                                                                       momentum quantum number with P = (-1) l . The total parity of a multiparticle state is the
The configuration of a real nuclide (which of course has both neutrons and protons) describes          product of the parities of the individual particles. A pair of particles with the same l will
the filling of its energy levels (sub-shells), for protons and for neutrons, in order, with the        therefore always have a combined parity of +1. The pairing hypothesis then tells us that the
notation (nlj)k for each sub-shell, where k is the occupancy of the given sub-shell. Sometimes,        total parity of a nucleus is found from the product of the parities of the last proton and the last
                                                                                                       neutron. So we can predict the parity of any nuclide, including the odd/odd ones and these
for brevity, the completely filled sub-shells are not listed, and if the highest sub-shell is nearly   predictions are in agreement with experiment.
filled, k can be given as a negative number, indicating how far from being filled that sub-shell
is. Using the ordering diagram above, and remembering that the maximum occupancy of each               The shell model can also be used to make predictions about nuclear magnetic (dipole)
sub-shell is 2j+1, we predict a configuration for 17O ( N = 9) of
                                                    8                                                  moments. Unless the nuclear spin is zero, we expect nuclei to have magnetic moments, since
                                                                                                       both the proton and the neutron have intrinsic magnetic moments, and the proton is
                          (1s1/2 )2 (1p3/2 )4 (1p1/2 )2         for the protons                        electrically charged, so it can produce a magnetic moment when it has orbital motion. Using a
and                                                                                                    notation similar to that used in atomic physics, we can write the nuclear magnetic moment as
                     (1s1/2 )2 (1p3/2 )4 (1p1/2 )2 (1d5/2 ) 1      for the neutrons                                                                m = g j j mN
Notice that all the proton sub-shells are filled, and that all the neutrons are in filled sub-shells   where m N is the nuclear magneton given by
except for the last one, which is in a sub-shell on its own. Most of the ground state properties
of 17O ( N = 9) can therefore be found from just stating the neutron configuration as (1d5 / 2 ) 1.
    8                                                                                                                                            m N = eh 2 M p
7.3 Spins, parities and magnetic moments in the shell model                                            g j is the Landé g-factor, j is the nuclear spin quantum number, and Mp is the proton mass. For
The nuclear shell model can be used to make predictions about the spins of ground states. A
                                                                                                       brevity we can write simply m = g j j magnetons.
filled sub-shell must have zero total angular momentum, because j is always an integer-plus-a-
half, so the occupancy of the sub-shell, 2j+1, is always an even number. This means that in a
filled sub-shell, for each nucleon of a given m j (= jz ) there is another having the opposite         We will find that the shell model does not give very accurate predictions for magnetic
                                                                                                       moments, even for the even-odd nuclei when there is only a single unpaired nucleon in the
 m j , so that the pair have a combined m j of zero, and so the complete sub-shell will also have
                                                                                                       ground state. We will not therefore consider at all the much more problematic case of the odd-
zero m j . Since this is true whatever axis we choose for z, the total angular momentum must           odd nuclei having an unpaired proton and an unpaired neutron.
also be zero. Since magic number nuclides have closed sub-shells, such nuclides are predicted
to have zero contribution to the nuclear spin from the neutrons or protons or both, whichever          For the even-odd nuclei, we would expect all the paired nucleons to contribute no net
are of magic number. Hence magic-Z/magic-N nuclei are predicted to have zero nuclear spin.             magnetic moment, for the same reason that they do not contribute to the nuclear spin.
This is indeed found to be the case experimentally.                                                    Predicting the nuclear magnetic moment is then a matter of finding the correct way to
                                                     7.7                                                                                               7.8
combine orbital and intrinsic components of magnetic moment of the single unpaired nucleon.                                        17
                                                                                                     As an example, consider        8 O,   which we found to have a ground state configuration of
We need to combine the spin component of the moment, gs s , with the orbital component,
 gl l (where gs and gl are the g-factors for spin and orbital angular momentum.) to give the
total moment g j j . The general formula for doing this is
                                                                                                                                  (1s1/2 )2 (1p3/2 )4 (1p1/2 )2         for the protons
                      j ( j + 1) + l(l + 1) - s( s + 1)      j ( j + 1) - l(l + 1) + s( s + 1)                               (1s1/2 )2 (1p3/2 )4 (1p1/2 )2 (1d5/2 ) 1      for the neutrons
               gj =                                     gl +                                   gs
                                  2 j ( j + 1)                           2 j ( j + 1)
                                                                                                     All the proton sub-shells are filled, and all the neutrons are in filled sub-shells except for the
which simplifies considerably because we always have j = l ± 1 / 2 . Thus                            last one, which is in a sub-shell on its own. There are three possibilities to consider for the
                                                                                                     first excited state:
                                   j g j = gl l + gs 2     for j = l + 1 2
                                                                                                     1     Promote one of the 1 p1/ 2 protons to 1d5 / 2 , giving a configuration of (1 p1/ 2 )                (1d5/2 )1,
                                                                                                           where the superscript –1 means that the shell is one particle short of being filled.
                                     Ê    1 ˆ        Ê 1 ˆ
                         jg j = gl j Á1 +   ˜ - gs j Á      ˜        for j = l - 1 2                                                                                                               -1
                                     Ë 2l +1¯        Ë 2l +1¯                                        2     Promote one of the 1 p1/ 2 neutrons to 1d5 / 2 , giving a configuration of (1 p1/ 2 )        (1d5/2 )2 .
                                                                                                     3     Promote the 1d5 / 2 neutron to the next level, which is probably 2 s1/ 2 (or the nearby 1d3 / 2 ),
Since gl = 1 for a proton and 0 for a neutron, and gs is approximately +5.6 for the proton and                                                  1              1
                                                                                                           giving a configuration of (1s1/ 2 ) or (1d3 / 2 ) .
-3.8 for the neutron, we find
                              j g proton = l + 5.6 ¥ = j + 2.8 for j = l + 1 2                       Following the diagram of Fig.7.3, the third of these possibilities would correspond to the
                                                     2                                               smallest energy shift, so it should be favoured over the others. The next excited state might
                                Ê        1 ˆ             Ê 1 ˆ         2.3                           involve moving the last neutron up a further level to 1d3 / 2 , or putting it back where it was and
                j g proton = j Á1 +          ˜ - 5.6 ¥ j Á      ˜ = 1-       for j = l - 1 2
                                Ë 2l +1¯                 Ë 2l +1¯      j +1                          adopting configurations (1) or (2). Option (2) is favoured over (1) because it keeps the excited
                                                      1                                              neutron paired with another, which should have a slightly lower energy than creating two
                                 j gneutron = -3.8 ¥ = -1.9 for j = l + 1 2                          unpaired protons. When comparing these predictions with the observed excited levels it is
                                                                                                     found that the expected excited states do exist, but not necessarily in precisely the order
                                                  Ê 1 ˆ 1.9 j
                            j gneutron = 3.8 ¥ j Á        ˜=         for j = l - 1 2                 anticipated.
                                                  Ë 2l +1¯ j +1
                                                                                                     The shell model has many limitations, most of which can be traced to its fundamental
For a given j, the measured moments usually lie somewhere between the j = l - 1 2 and the            assumption that the nucleons move independently of one another in a spherically symmetric
 j = l + 1 2 values, but beyond that, the model does not predict the moments accurately, except      potential. The latter, for example, is only true for nuclei that are close to having doubly filled
for nuclei with small numbers of nucleons which are close to magic values.                           magnetic shells and predicts zero quadruple moments, whereas in practice nuclei are
                                                                                                     deformed and quadruple moments are often substantial.
Why should the shell model work so well when predicting nuclear spins and parities, but be
poor for magnetic moments? There are several likely problem areas, including the possibility         7.5 Collective Model
that protons and neutrons inside nuclei may have effective intrinsic magnetic moments that           The shell model is based upon the idea that the constituent parts of a nucleus move
are different to their free-particle values, because of their very close proximity to one another.   independently. The liquid drop model implies just the opposite, since in a drop of
                                                                                                     incompressible liquid, the motion of any constituent part is correlated with the motion of all
7.4 Excited states in the shell model                                                                the neighbouring pairs. This emphasizes that models in physics have a limited range of
In principle, the shell model’s energy level structure can be used to predict nuclear states away    applicability and may be unsuitable if applied to a different set of phenomena. As knowledge
from the ground state, i.e. the excited states. This works quite well for the first one or two       evolves, it is natural to try and incorporate more phenomena by modifying the model to
excited states when there is only one possible configuration of the nucleus. But for higher          become more general, until (hopefully) we have a model with firm theoretical underpinning
excited states the spectrum becomes very complicated because several nucleons can be                 that describes a very wide range of phenomena, i.e. a theory. A step in this direction in nuclear
excited simultaneously into a superposition of many different configurations to produce a            physics is the collective model which uses the ideas of both the shell and liquid drop models.
given nuclear spin and parity.
                                                                                                     The collective model assumes that the nucleons in unfilled subshells move independently in
When trying to predict the first one or two excited states using a filling diagram like Fig.7.3,     the nuclear potential produced by the core of filled subshells, as in the shell model. However,
we are looking for the configuration that is nearest to the ground state configuration. This will    this potential is not the static, spherically symmetric potential used in the shell model, but is a
normally involve either moving an unpaired nucleon to the next highest level, or moving a            potential capable of undergoing deformations in shape. These deformations represent the
nucleon from the sub-shell below the unpaired nucleon up one level to pair with it. Thus it is       collective (correlated) motion of the nucleons in the core of the nucleus that are associated
necessary to consider levels just above and below the last nucleons (protons and neutrons).          with the liquid drop model. I will not discuss this model further, except to say that it can
                                                     7.9                                                                                                    7.10
explain some properties of nuclei (e.g. electric quadruple moments) that other models cannot.         called the matrix element because it is one element of a matrix whose elements are all the
A summary of the features of various nuclear models is given in Table 7.1                             possible transitions to different final states of the system) and n ( E ) is the density of states,
                                                                                                      i.e. number of quantum states available to the final system per unit interval of total energy.
             Table 7.1 Nuclear models and the ground state properties of nuclei                       The density-of-states factor can be calculated from purely kinematical factors, such as
                                                                                                      energies, momenta, masses, and spins where appropriate. (We did this when discussing the
                                                                                                      relationship between the scattering amplitude and cross-sections and also in the Fermi gas
                                                                                                      model) The dynamics of the process is contained in the matrix element.
                                                                                                      The matrix element can in general be written in terms of one of five basic classes of Lorentz
                                                                                                      invariant interaction operators, O :
                                                                                                                                            M = Ú Y f* ( gO) Yi d V
                                                                                                      where Yf and Y i are total wave-functions for the final and initial states, respectively, g is a
                                                                                                      dimensionless coupling constant, and the integral is over three-dimensional space
                                                                                                      (V = volume). The five categories are called scalar (“S”), pseudo-scalar (“P”), vector (“V”),
                                                                                                      axial-vector (“A”), and tensor (“T”), the names having their origin in the mathematical
                                                                                                      transformation properties of the operators. (We have met the V and A forms previously in
                                                                                                      Section 6 on the electroweak interaction.) The main difference between them is the effect on
                                                                                                      the spin states of the particles. When there are no spins involved, and at low energies, (g O ) is
                                                                                                      simply the interaction potential: that part of the potential that is responsible for the change of
                                                                                                      state of the system.
                                                                                                      Fermi guessed that O would be of the vector type, since the electromagnetic interaction is a
7.6   b - Decay : theory                                                                              vector interaction, i.e. it is transmitted by a spin-1 particle – the photon. We have seen from
                                                                                                      earlier lectures that we now know that the weak interaction violates parity conservation and is
(a) Fermi theory                                                                                      correctly written as a mixture of both vector and axial-vector interactions, but so long as we
The first successful theory of nuclear beta-decay was proposed in the 1930s by Fermi, long            are not concerned with the spins of the particles, this doesn't make much difference, and we
before the W and Z bosons were known and the quark model formulated. He therefore had to              can think of the matrix element in terms of a classical weak interaction potential, like the
construct a theory based on very general principles, working by analogy with the quantum              Yukawa potential. Applying a bit of modern insight, we can consider that the potential is of
theory of electromagnetic processes (QED).                                                            extremely short range (because of the large mass of the W boson) in which case we have seen
                                                                                                      that we can approximate the interaction by a point-like interaction and the matrix element
The general equation for electron b -decay is                                                         becomes simply a constant, written as
                                      A       A
                                      Z X Æ Z +1Y   + e- + ne                                                                                      M=      .
which we have met in earlier lectures. There we interpreted this reaction as the decay of a
                                                                                                      V is the arbitrary volume, which is used to normalise the wave-functions. (It will eventually
bound neutron, i.e. n Æ p + e - + n e and later we gave the quark interpretation of this decay.       cancel out with a factor coming from the density of states term.) GF is called the Fermi
In general, it is possible for the internal state of the nucleus to change in other ways during the   coupling constant and has dimensions [energy][length]3. In nuclear theory, Fermi's coupling
transition, but we will simplify matters by considering just the basic neutron decay process.         constant GF is taken to be a universal constant and with appropriate corrections for changes
                                                                                                      of the nuclear state during beta decay, experimental results are consistent with the theory
We have also met Fermi’s Second Golden Rule, which enables transition rates to be                     under this assumption. But the theory goes beyond nuclear beta decay, and can be applied to
calculated provided the interaction is relatively weak. (This is because it is based on               any process mediated by the W-boson — provided the energy is not too great. Thus, the best
perturbation theory.) We will write Fermi’s Rule as                                                   process to determine the value of GF is one not complicated by hadronic (nuclear) effects and
                                             2p                                                       from purely muon decay one can deduce that the value of GF is about 90 eV fm3. It is often
                                        w=      M n( E )                                              quoted in the form GF ( hc ) 3 = 1.166 x10 -5 GeV -2 .
where w is the transition rate (probability per unit time), M is the transition amplitude (also
                                              7.11                                                                                                   7.12
(b) Electron momentum distribution                                                                                                                                           2
                                                                                                                             d w GF pe pn GF pe En GF pe ( E - E e )
                                                                                                                                     2 2 2       2 2 2        2 2
We see that the transition rate (i.e. beta-decay lifetime) depends essentially on kinematical                                     =           =             =
factors arising through the density-of-states factor, n ( E ) . To simplify the evaluation of this
                                                                                                                             d pe   2p 3 h 7c   2p 3 h 7c 3    2p 3 h 7c 3
factor, we consider the neutron and proton to be “heavy”, so that they have negligible kinetic
energy, and all of the energy released in the decay process goes into creating the electron and      This expression gives rise to a bell-shaped electron momentum distribution, which rises from
neutrino and in giving them kinetic energy. Thus we write                                            zero at zero momentum, reaches a peak and falls to zero again at an electron energy equal to
                                                                                                     E, as illustrated in the curve labelled Z = 0 in Fig.7.4. Studying the precise shape of the
                                                   E = E e + En                                      distribution near its upper end-point is one way in principle of finding a value for the
                                                                                                     antineutrino mass. If the mass is zero mass, then the gradient of the curve approaches zero at
where E e is the total (relativistic) energy of the electron, En is the total energy of the          the end-point, whereas any non-zero value results in an end-point that falls to zero with an
                                                                                                     asymptotically infinite gradient. (See Fig.7.4 – note that the insert to this figure is not
neutrino, and E is the total energy released [= (Dm)c2, if Dm is the neutron-proton mass
                                                                                                     accurately drawn.) We return to this below.
difference, or the change in mass of the decaying nucleus].
The transition rate, w , can be measured as a function of the electron momentum, so we need
to obtain an expression for the spectrum of beta-decay electrons. Thus we will fix E e and find
the differential transition rate for decays where the electron has an energy in the range E e to
 E e + dE e . From Fermi’s Rule, this is
                                          2p  2
                                dw =         M nn ( E - E e ) n e ( E e ) dE e
where n e and nn are the density of states factors for the electron and neutrino, respectively.
These may be obtained from our previous result:
                                       n ( pe ) dpe =            4p pe 2 dpe
                                                         (2ph) 3
with a similar expression for nn , by changing variables using dp dE = E pc 2 , so that                       Fig.7.4 Predicted electron spectrum: Z = 0 , without Fermi screening factor;
                                                                                                                               b ± , spectra with Fermi screening factor
                                      n ( E e ) dE e =               pe E e dE e
                                                         (2ph) 3 c 2                                 There are several factors that we have ignored or over-simplified in getting this momentum
                                                                                                     distribution. The principal ones are to do with the possible changes of nuclear spin of the
with a similar expression for n ( En ) . Using these in the expression for dw and setting            decaying nucleus, and the electric force acting between the beta particle (electron or positron)
M = GF V , gives                                                                                     and the nucleus. When the electron/antineutrino carry away a combined angular momentum of
                                                                                                     0 or 1, the above treatment is essentially correct: these are the so-called “allowed transitions”.
                                         dw     GF 2
                                             =           p E p E                                     However, the nucleus may change its spin by more than 1 unit, and then the simplified short-
                                         dE e 2p 3 h 7c 4 e e n n                                    range-potential approach to the matrix element is inaccurate. The decay rate in these cases is
where in general                                                                                     generally suppressed, although not completely forbidden, despite these being traditionally
                                                                                                     known as the “forbidden transitions”.
                             pn c =       2    2
                                         En - mn c 4 =          ( E - E e )2 - mn2 c 4
                                                                                                     The electric potential between the positive nucleus and a positive beta particle will cause a
                                                                                                     shift of the low end of its momentum spectrum to the right, since it is propelled away by
Finally, it is useful to write this as
                                                                                                     electrostatic repulsion. Conversely, the low end of the negative beta spectrum is shifted to the
                                                                                                     left. (See Fig.7.4.) The precise form of these effects is difficult to calculate, and requires
                                dw dE e dw     GF 2
                                   =        =           p 2p E                                       quantum mechanics, but the results are published in tables of a factor F ( Z, E e ) , called the
                                dpe dpe dE e 2p 3 h 7c 2 e n n                                       Fermi screening factor, to be applied to the basic beta spectrum.
If we take the antineutrino to be precisely massless, then pn = En c and we have                     (c) Kurie plots and the neutrino mass
                                                                                                     The usual way of experimentally testing the form of the electron momentum spectrum given
                                                                                                     by the Fermi theory is by means of a Kurie plot. From the above equations, with the Fermi
                                                         7.13                                                                                      7.14
screening factor included, we have
                                 d w F ( Z, E e ) GF pe ( E - E e )
                                                   2    2               2
                                 d pe           2p 3 h 7c 3
which we can write as
                                        Ê dw ˆ        1
                             H ( Ee ) ∫ Á      ˜ 2              = E - Ee
                                        Ë d pe ¯ pe K ( Z, pe )
where K ( Z, pe ) includes F ( Z, E e ) and all the fixed constants. A plot of the left-hand-side of
this equation — using the measured dw dpe and pe and the calculated K ( Z, pe ) — against
the electron energy E e should then give a straight line with an intercept of E. An example is                 Fig.7.6 Schematic Kurie plot. If the neutrino mass is not zero, the straight line
shown in Fig.7.5.                                                                                                 curves near the end point and cuts the axis vertically at E 0 = E 0 - mn c 2 .
                                                                                                       In order to measure the neutrino mass to reasonable accuracy, it is necessary to use a nucleus
                                                                                                       where the non-zero mass has a maximum effect, i.e. the maximum energy release E = E 0
                                                                                                       should only be a few kev. Also at low energies atomic effects have to be taken into account,
                                                                                                       so the initial and final atomic states must be very well understood. The most suitable case is
                                                                                                       the decay of tritium
                                                                                                                                                   H Æ 3He + e - + n e
                                                                                                       where E 0 = 18.6 keV . Since the counting rate near E 0 is vanishingly small, the experiment is
                                                                                                       extremely difficult. In practice, the above formula is fitted to data close to the end point of the
                                                                                                       spectrum and extrapolated to E 0 . The best experiments are consistent with a zero neutrino
                                                                                                       mass, but when experimental and theoretical uncertainties are taken into account, an upper
                                                                                                       limit of 3 eV c 2 is quoted.
                                    Electron kinetic energy in keV
                            Fig.7.5 Kurie plot for the decay 36Cl .
                   (The y-axis is proportional to the function H ( E e ) above.)
If the neutrino mass is not exactly zero then it is straightforward to repeat the above derivation
and to show that the left-hand side of the Kurie plot is proportional to
                                   ( E - E e ) ( E - E e )2 - mn 2c 4
This will produce a very small deviation from linearity close to the end point of the spectrum
as shown schematically in Fig.7.6. (n.b. the y-axis of this graph is actually the function
H ( E e ) defined earlier.)
                                                7.15                                                                                                     7.16

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